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RideAnS [48]
3 years ago
11

What atoms make up a carboxyl group?

Chemistry
2 answers:
sineoko [7]3 years ago
8 0

Answer:

Carbon , oxygen and hydrogen.

Explanation:

A carboxyl group is COOH.

statuscvo [17]3 years ago
7 0

Answer:

The answer to your question is below

Explanation:

Carboxyl group is the characteristic group of carboxylic acids.

It is composed by 1 atom of carbon, 2 atoms of oxygen and 1 atom of hydrogen.

Its structure is   - COOH

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Where did the longest tsunami form?
max2010maxim [7]
July 9, 1958: in Lituya Bay<span>, Alaska was caused by a landslide triggered by an 8.3 magnitude earthquake.</span>
8 0
2 years ago
Read 2 more answers
Write the complete ionic equations, spectator ions and net ionic equation for the following.
gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq)  →

Chemical equation:

HgCl₂ (aq) + KI (aq)  → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq)  → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq)  + Cl₂²⁻  (aq) + 2K⁺(aq) + 2I⁻ (aq)  →  HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq)  + 2I⁻ (aq) →   HgI₂ (s)

The Cl⁻ ((aq)  and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

 

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0
3 years ago
The scientific notation 2 x 10-2 has what value? <br> 0.02<br> 0.002<br> 0.2<br> 0.0002
Marta_Voda [28]
Since the exponent is negative, you move the decimal (2.0) to the left two spots leaving you with .02
7 0
2 years ago
Read 2 more answers
What is the concentration of an naoh solution that requires 15.0 ml of a 0.750 m h2so4 solution to neutralize 17.5 ml of naoh?
e-lub [12.9K]

So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
                      2NaOH  +  H₂SO₄   →  Na₂SO₄<span>  +  2H₂O
</span>
So now it's time to identify what reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)

If 1000 ml  of H₂SO₄ contain 0.750 mol   [0.750 M is the amount of moles in 
                                                                       1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol     [15 ml is the amount of the acid that                                                                                  took part in the reaction]
    
⇒  x  =  \frac{15ml    *    0.750 mol}{1000 ml}
       
         = 0.01125 mol

Mole ratio of NaOH  to  H₂SO₄  can be obtained from the balanced equation
                 0 2NaOH  +  1H₂SO₄   →  Na₂SO₄  +  2H₂O

    mole ratio of   NaOH  to  H₂SO₄  is 2 : 1

∴ if mole of of H₂SO₄   =  0.01125 mol
   then moles of NaOH = (0.01125 mol) × 2
                                      = 0.0225 mol

If 17.5 ml of NaOH contain 0.0225 mol      [this was given in the question]
then let 1000 ml of NaOH contain  x

⇒ x  =  \frac{1000ml   * 0.0225 mol}{17.5 ml}
       
       = 1.286 mol

∴ concentration of NaOH is 1.286 mol/L 

8 0
3 years ago
Which of the following buffers will be most effective at pH 9.25? Group of answer choices a mixture of 1.0 M HC2H3O2 and 1.0 M N
ludmilkaskok [199]

Answer:

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

Explanation:

Step 1: Data given

pH of a buffer = pKa + log ([A-]/[Ha])

a mixture of 1.0 M HC2H3O2 and 1.0 M NaC2H3O2 (Ka for acetic acid = 1.8 x 10-5)

pH = -log( 1.8 * 10^-5) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 4.74

a mixture of 1.0 M NaCN and 1.0 M KCN (Ka for HCN = 4.9 x 10-10)

pH = -log( 4.9 * 10^-10) + log (1/1)

pH = -log( 1.8 * 10^-5)

pH = 9.30

a mixture of 1.0 M HCl and 1.0 M NaCl

The solution made from NaCl and HCl will NOT act as a buffer.

HCl is a strong acid while NaCl is salt of strong acid and strong base which do not from buffer solutions hence due to HCl PH is less than 7.

a mixture of 1.0 M NH3 and 1.0 M NH4Cl (Kb for ammonia = 1.76 x 10^-5)

Ka * Kb = 1*10^-14

Ka = 10^-14 / 1.76*10^-5

Ka = 5.68*10^-10

pH = -log( 5.68*10^-10) + log (1/1)

pH = -log( 5.68*10^-10)

pH = 9.25

The most effective buffer at pH 9.25 will be  a mixture of 1.0 M NH3 and 1.0 M NH4Cl

8 0
3 years ago
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